If $\mathbb C=(0<1)$ is the walking arrow, then the question of composability reduces to the question of whether, given that the two small hexagons below commute, the outer hexagon commutes:
It’s straightforward enough to arrange that $G$ makes it easy for $\alpha,\beta$ to be dinatural while $F$ and $H$ are more resistant for $\alpha \beta$ to be so. In particular, let's let $G(1,0)$ be initial in $\mathbb D$ and $G(0,1)$ be terminal so that $\alpha$ and $\beta$ will be dinatural no matter what other choices we make. Let $x$ be some object of $\mathbb D$ admitting a nontrivial endomorphism $\varphi$ and let $F,H,$ and the diagonal of $G$ be constant at $x.$ Then the outer hexagon is commutative if and only if $\alpha_0\beta_0=\alpha_1\beta_1,$ which we can prevent by picking exactly one of those four morphisms to be $\varphi.$
So for a completely specific example, we see it would work to take $\mathbb D$ a category freely generated by objects $0,1,2$ subject to the conditions that $0$ is initial, $1$ is terminal, and $\mathbb D(2,2)$ is, say, $\mathbb Z/2\mathbb Z.$ I think there are only eight morphisms here, and it seems like it would be hard to go much smaller (though the other answer currently visible does a bit better.)